A certain mass of a substance- when dissolved in 100 g C6H6- lowers the freezing point by 1-28-C- The same mass of solute dissolved in 100 g water lowers the freezing point by 1-40-C- If the substance has normal molecular weight in benzene and is completely ionized in water- into how many ions does it dissociate in water-Given- Kf for H2O and C6H6 are 1-86 and 5-12 K kg mol-1

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Standard X

Chemistry

Solubility

A certain mas...

Question

A certain mass of a substance, when dissolved in $100 g C_{6} H_{6}$ , lowers the freezing point by ${1.28}^{\circ} C$ . The same mass of solute dissolved in $100 g$ water lowers the freezing point by ${1.40}^{\circ} C$ . If the substance has normal molecular weight in benzene and is completely ionized in water, into how many ions does it dissociate in water?
(Given: $K_{f}$ for $H_{2} O$ and $C_{6} H_{6}$ are $1.86$ and $5.12 K k g m o l^{- 1}$ )

1 ions

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3 ions

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2 ions

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will not dissociate

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Solution

The correct option is C 3 ions
In benzene, substance lowers the freezing point by ${1.28}^{o}$ C.

$1.28 = 1 \times K_{f} \times$ molality

$1.28 = 1 \times 5.12 \times$ molality

Molality $= 0.25$

In water , substance lowers the freezing point by ${1.40}^{o}$ C

$1.40 = 1 \times K_{f} \times$ molality $\times i$

$1.40 = 1 \times 1.86 \times 0.25 \times i$

$i = 3$

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The correct option is C 3 ionsIn benzene, substance lowers the freezing point by 1.28o C.1.28=1×Kf×molality1.28=1×5.12×molalityMolality =0.25In water , substance lowers the freezing point by 1.40o C1.40=1×Kf× molality ×i1.40=1×1.86×0.25×ii=3